Related Concepts: 1.01 Signals, Systems & Singularity Functions | 1.02 Elementary Operations on Signals | 1.03 Signal Classification I — Theory & Periodicity

1.04 Signal Classification II — Symmetry, Energy & Power

Overview

Signals are further classified mathematically based on time boundaries, symmetry across axes, and physical energy/power content.

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1. Causal, Anti-causal, and Non-causal Signals

This classification is determined entirely by where the signal exists relative to the $t=0$ boundary on the horizontal axis.

• Causal Signal: Only exists in positive time. It starts at or after $t=0$.
  • Mathematical Condition: $x(t)=0$ for $t<0$.
  • Tip: Any signal multiplied by $u(t)$ is automatically forced to be causal.
• Anti-causal Signal: Only exists in negative time. It ends at or before $t=0$.
  • Mathematical Condition: $x(t)=0$ for $t>0$.
  • Tip: Any signal multiplied by $u(-t)$ is anti-causal.
• Non-causal Signal: Exists on both sides of the $t=0$ boundary (extends into both positive and negative time).

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Graphical Examples

Causal Signal ($e^{-2t}u(t)$)

left = -3; right = 4
top = 1.5; bottom = -0.5
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y = exp(-2*x) | x >= 0 | BLUE | SOLID
y = 0 | x < 0 | BLUE | SOLID
(0, 1) | label:Causal e^(-2t)u(t) | BLUE

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Anti-causal Signal ($e^{2t}u(-t)$)

left = -4; right = 3
top = 1.5; bottom = -0.5
---
y = exp(2*x) | x <= 0 | RED | SOLID
y = 0 | x > 0 | RED | SOLID
(0, 1) | label:Anti-causal e^(2t)u(-t) | RED

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Non-causal Signal ($e^{-t^2}$)

left = -3; right = 3
top = 1.5; bottom = -0.5
---
y = exp(-x^2) | BLUE | SOLID
(0, 1) | label:Non-causal e^(-t^2) | BLUE

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Past Year Questions (PYQs)

• [PYQ 2025, 2020]: State whether the following signals are causal, anticausal or noncausal. (i) $x(t)=e^{-2t}u(t)$; (ii) $x(t)=tu(t)$. (04 Marks)
  • Solution: Both signals are multiplied by $u(t)$. Since $u(t)=0$ for $t<0$, both $x(t)=0$ for $t<0$, making them strictly Causal signals.

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2. Even and Odd Signals (Symmetry)

Any arbitrary signal $x(t)$ can be decomposed into a sum of an even component $x_e(t)$ and an odd component $x_o(t)$.

• Even Signal: Symmetrical across the vertical $y$-axis. $x(t)=x(-t)$
• Odd Signal: Anti-symmetrical across the origin. $x(t)=-x(-t)$
• Decomposition Formulas: $x_e(t)=\frac{x(t)+x(-t)}{2}$ $x_o(t)=\frac{x(t)-x(-t)}{2}$ $x(t)=x_e(t)+x_o(t)$

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Graphical Decomposition of $u(t)$

Even Component of Step Function: $u_e(t)=1/2$

left = -4; right = 4
top = 1; bottom = -1
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y = 0.5 | BLUE | SOLID
(0, 0.5) | label:ue(t) = 1/2 | BLUE

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Odd Component of Step Function: $u_o(t)=\frac{1}{2}\text{sgn}(t)$

left = -4; right = 4
top = 1; bottom = -1
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y = 0.5 | x > 0 | RED | SOLID
y = -0.5 | x < 0 | RED | SOLID
(0, 0) | label:uo(t) = 1/2 sgn(t) | RED
(0, 0.5) | open | RED
(0, -0.5) | open | RED

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Even/Odd Energy Component Proof

Symmetrical Definite Integral Energy Identity

Question: If $x_e(t)$ and $x_o(t)$ are the even and odd components of $x(t)$ respectively, prove that:

${\int }_{-\infty }^{\infty }{x}^2(t)dt={\int }_{-\infty }^{\infty }{x}_e^2(t)dt+{\int }_{-\infty }^{\infty }{x}_o^2(t)dt$ (PYQ 2024, 2023, 2021)

Proof:

1. Express $x(t)$ as the sum of its components: $x(t)=x_e(t)+x_o(t)$
2. Square both sides and integrate over all time: ${\int }_{-\infty }^{\infty }{x}^2(t)dt={\int }_{-\infty }^{\infty }{\left[x_e(t)+x_o(t)\right]}^{2}dt$ ${\int }_{-\infty }^{\infty }{x}^2(t)dt={\int }_{-\infty }^{\infty }{x}_e^2(t)dt+{\int }_{-\infty }^{\infty }{x}_o^2(t)dt+2{\int }_{-\infty }^{\infty }{x}_e(t)x_o(t)dt$
3. Analyze the cross-product term: $x_e(t)\cdot x_o(t)$.
   • The product of an even function and an odd function is always an odd function.
   • The definite integral of any odd function over symmetrical limits ($-\infty$ to $\infty$) is mathematically equal to zero. $2{\int }_{-\infty }^{\infty }{x}_e(t)x_o(t)dt=0$
4. Therefore, the expression simplifies to: ${\int }_{-\infty }^{\infty }{x}^2(t)dt={\int }_{-\infty }^{\infty }{x}_e^2(t)dt+{\int }_{-\infty }^{\infty }{x}_o^2(t)dt■$

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3. Energy and Power Signals

This classification measures whether a signal eventually decays to zero (Energy) or continues indefinitely with steady amplitude (Power).

3.1 Mathematical Definitions

• Total Energy ($E$):
  • Continuous-Time: $E={\int }_{-\infty }^{\infty }|x(t){|}^{2}dt$
  • Discrete-Time: $E={\sum }_{n=-\infty }^{\infty }|x[n]{|}^2$
• Average Power ($P$):
  • Continuous-Time: $P={\lim }_{T\to \infty }\frac{1}{T}{\int }_{-T/2}^{T/2}|x(t){|}^{2}dt$
  • Discrete-Time: $P={\lim }_{N\to \infty }\frac{1}{2N+1}{\sum }_{n=-N}^N|x[n]{|}^2$

graph TD
    Start["Calculate E and P"] --> CheckE{"Is E finite?"}
    CheckE -->| "Yes: 0 < E < ∞" | Energy["Energy Signal: P = 0"]
    CheckE -->| "No: E = ∞" | CheckP{"Is P finite?"}
    CheckP -->| "Yes: 0 < P < ∞" | Power["Power Signal: E = ∞"]
    CheckP -->| "No: P = ∞" | Neither["Neither Energy nor Power"]

The Golden Classification Rules

1. Energy Signal: $0<E<\infty$ and $P=0$. (Most transient, finite-duration, or exponentially decaying pulses).
2. Power Signal: $0<P<\infty$ and $E=\infty$. (All periodic signals, and infinite-duration steps).
3. A signal cannot be both.
4. If a signal grows infinitely (e.g., $e^{at}$, $t\cdot u(t)$), then $E=\infty$ and $P=\infty$, so it is neither.

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3.2 Numerical Trap Solutions

Trap 1: The Discrete Zero-Energy Identity

Question: Determine whether the discrete signal $x[n]$ is an energy or power signal, and find its value: $x[n]=\{\begin{array}{ll}\sin (\pi n),& n\ge 0\\ 0,& \text{otherwise}\end{array}$ (PYQ 2025)

Solution:

1. Note that $n$ is strictly an integer index.
2. For any integer $n$, $\sin (\pi n)=0$ (e.g., $\sin (0)=0,\sin (\pi )=0,\sin (2\pi )=0$).
3. Therefore, $x[n]=0$ for all $n$.
4. $E=0\text{and}P=0$
5. Conclusion: It is technically neither an energy nor a power signal (it is a trivial zero signal).

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Trap 2: Discrete Cosine Power Signal

Question: Determine whether the signal $x[n]$ is an energy or power signal, and find its value: $x[n]=\{\begin{array}{ll}\cos (\pi n),& n\ge 0\\ 0,& \text{otherwise}\end{array}$ (PYQ 2023, 2019)

Solution:

1. For integers $n$, $\cos (\pi n)=(-1{)}^n$.
2. Calculate Energy: $E={\sum }_{n=0}^{\infty }|(-1{)}^n{|}^2={\sum }_{n=0}^{\infty }1=\infty ⟹\text{Not an Energy Signal}$
3. Calculate Power: $P={\lim }_{N\to \infty }\frac{1}{2N+1}{\sum }_{n=0}^N|(-1{)}^n{|}^2={\lim }_{N\to \infty }\frac{N+1}{2N+1}$ Using L’Hôpital’s Rule or dividing by $N$: $P=\frac{1}{2}\text{ Watts}$
4. Conclusion: Since $E=\infty$ and $P=0.5\text{ W}$, it is a Power Signal.

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Trap 3: The Continuous-Time Step

Question: Determine the power and energy of the continuous-time step function $x(t)=u(t)$. (2022)

Solution:

1. Calculate Energy: $E={\int }_{-\infty }^{\infty }|u(t){|}^{2}dt={\int }_0^{\infty }1dt=\infty$
2. Calculate Power: $P={\lim }_{T\to \infty }\frac{1}{T}{\int }_{-T/2}^{T/2}|u(t){|}^{2}dt={\lim }_{T\to \infty }\frac{1}{T}{\int }_0^{T/2}1dt={\lim }_{T\to \infty }\frac{T/2}{T}=\frac{1}{2}\text{ Watts}$
3. Conclusion: It is a Power Signal with $P=0.5\text{ Watts}$.

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Trap 4: Bounded Discrete Range

Question: Determine whether the signal $x[n]$ is an energy or power signal, and find its value: $x[n]=\{\begin{array}{ll}\cos (\pi n),& -4\le n\le 4\\ 0,& \text{otherwise}\end{array}$ (PYQ 2018)

Solution:

• Exam Warning: Any bounded, finite-duration signal is always an Energy Signal.
• Calculate Energy: $E={\sum }_{n=-4}^4|\cos (\pi n){|}^2={\sum }_{n=-4}^4|(-1{)}^n{|}^2={\sum }_{n=-4}^{4}1=9\text{ Joules}$
• Average Power $P=0$.
• Conclusion: It is an Energy Signal with $E=9\text{ Joules}$.

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Additional Past Year Questions (PYQs)

• [PYQ 2020]: Find the odd and even components for each of the following signals: (i) $x(t)=u(t)$; (ii) $x(t)=e^{-\alpha t}u(t)$. (08 Marks)
  • Solution (i): $u_e(t)=\frac{u(t)+u(-t)}{2}=\frac{1}{2}$ (for $t\ne 0$); $u_o(t)=\frac{u(t)-u(-t)}{2}=\frac{1}{2}\text{sgn}(t)$.
  • Solution (ii): $x_e(t)=\frac{e^{-\alpha t}u(t)+e^{\alpha t}u(-t)}{2}$; $x_o(t)=\frac{e^{-\alpha t}u(t)-e^{\alpha t}u(-t)}{2}$.
• [PYQ 2018, 2015]: Find the even and odd components of the given signal $x(t)=\cos (20\pi t-5\pi )$ and sketch the signal. (10 Marks)
  • Hint: $\cos (20\pi t-5\pi )=-\cos (20\pi t)$. Since $\cos (-20\pi t)=\cos (20\pi t)$, this is an even function. Even component is $-\cos (20\pi t)$ and odd component is $0$.
• [PYQ 2019]: Determine the even and odd components of the signal $x(t)=\sin (\sqrt{2}\pi t)+\sin (15\pi t)$. (06 Marks)
  • Hint: Since $\sin (-\theta )=-\sin (\theta )$, this is a purely odd function. Even component is $0$ and odd component is $x(t)$.
• [PYQ 2017]: Determine whether $x[n]=\{\begin{array}{ll}n,& 0\le n<5\\ 10-n,& 5\le n\le 10\\ 0,& \text{otherwise}\end{array}$ is energy or power.
  • Answer: Bounded finite duration $⟹$ Energy signal. $E=0^2+1^2+2^2+3^2+4^2+5^2+4^2+3^2+2^2+1^2+0^2=85\text{ Joules}$.
• [PYQ 2016]: Consider a signal $x(t)=5\cos (\pi t)+\sin (5\pi t)$. Determine if it is an energy or power signal and find its value.
  • Answer: Power signal (sum of periodic components is periodic). $P=\frac{5^2}{2}+\frac{1^2}{2}=13\text{ Watts}$.
• [PYQ 2015]: Compute the energy of $x(t)=\sin (\sqrt{2}\pi t)+\sin (15\pi t)$.
  • Answer: Since sines run infinitely, $E=\infty$. It is a power signal with $P=\frac{1^2}{2}+\frac{1^2}{2}=1\text{ Watt}$.